4-internal forces beams _1 model (1)

7/23/2019 4-Internal Forces Beams _1 Model (1)

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S t r u c t u r a l A n a l y s i s I )

INTERNAL FORCES BEAMS

PART ( 4 )


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. #

. #

#

. #

. #

. #

INTERNAL FORCES

(M @ sec = F *dF)

@M@

PART (4)

2014

2014

4/1

Structural Analysis (I) Internal Forces/BEAMS

P

Section Member

Normal Force (N)

Shear Force (Q)

Bending Moment (M)

P

R1 R2 R2R1

Loads

Reactions

N

M

Q

. #

Normal Force Diagram (N.F.D)

Shear Force Diagram (S.F.D)Bending Moment Diagram (B.M.D)

#

Normal Force (N)

Shear Force (Q)

Bending Moment (M)

N

NPOSITIVE NORMAL FORCE

N

N

NN

N

NNEGATIVE NORMAL FORCECOMPRESSION

POSITIVE SHEAR FORCE

NEGATIVE SHEAR FORCE

Q

Q

Q

QQ

Q

Q

Q

POSITIVE BENDING MOMENT

M

TENSION

MMM

+ve

-ve

+ve

-ve

+ve-ve

-ve

+ve

Datum


3/15

@M@

PART (4)

2014

2014

4/2


EX Calculate Internal Forces at shown Sections then Draw I.F. Diagrams.

SEC 1

N = -8t (comp)

3t

8t1

Q = +3t ()

M = +3(0) =0 t.m

SEC 2

N = -8t (comp)

Q = +3t ()M = +3(3) =9 t.m

2

3t

8tN1

Q1

SEC 3N = -8t +8 =0t

Q = +3t () - 6t () = - 3t()

M = +3(3) = 9 t.m

N2

Q22

3t

8t 8t

6t

3

M2

SEC 4

N = 0t

Q = -3t()M = +3(6) - 6(3) = 0 t.m

N3

Q38t

6t4

3t

8t

Start from Left @ point a

REACTIONS

M a= 0.0

FY= 0.0

FX= 0.0 Xa= 8t

Ya= 3t

Yb= 3t

MM

NEGATIVE BENDING MOMENT

MM

a

Ya

Xa b

Yb

M

1 2

3

4

5

a

Ya

Xa b

Yb3t 3t

8t

10t3

4

8t

6t

9t.m1

2 3

4 5 6

/

/

/

6

7

8

9

wt/m`

#

+ve

-ve

Q

N

Datum

Left Right

+ve Sign summary


4/15

SEC 5

N = 0t

Q = -3t()

M = 0 + 9 = +9 t.m

N4

Q4

8t

6t

3t

8t9t.m 5

9t.m5

SEC 6

N = 0t

Q = -3t()M = 3*9-6*6+9=0 t.m

N5

Q5

CHECK Right part

3t

6

@M@

PART (4)

2014

2014

4/3


DRAWING

I.F.

SEC 1 2 3 4 5 6N

Q

M

-8t -8t 0t 0t 0t 0t

+3t +3t -3t -3t -3t -3t

0t.m +9t.m +9t.m 0t.m 0t.m+9t.m

a b

3t

10t3

4

8t

6t

9t.m1

2 3

4 5 6

3t

8t

8t

-

3t

3t

P=6t

9t.m9t.m

N.F.D

S.F.D

B.M.D

1cm=8t

1cm=3t

1cm=9t.m

-

+

M=9t.m

NONE

N = 0t Q = -3t()M = 0.0 t.m

M4

11

3

3

N=8t3

2

N.F.D & S.F.D

8t

6t

3t

8t9t.m 6

#

(

)

-1Datum

(

)

-2B.M.D. ( ( ( ) -3N.F.D , S.F.D &B.M.DN, Q &M Drop

S.F.D

.

Left / Right

Left Right

Drawing against force directio

Tension

Compression

Left Right

Drawing with force direction


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STANDARD CASES

@M@

PART (4)

2014

2014

4/4


P

S.F.D

B.M.D

P

P*L

+

10t

S.F.D

B.M.D

10t

30t.m

+

5t

S.F.D

B.M.D

5t

10t.m

+

P

S.F.D

B.M.D

P

P*L

+

Case (1)

Case (2)

Case (3)

S.F.D

B.M.D

NONE

10t.m

10t.m

S.F.D

B.M.D

NONE

M

M


6/15

8t

4t 4t

S.F.D

B.M.D

8*84 =16t.m

4t

4t-

+

6t

4t

S.F.D

B.M.D

6*2*46 =8t.m

-+

2t

4t

2t

@M@

PART (4)

2014

2014

4/5


P

P2

P2

S.F.D

B.M.D

PL4

P/2

P/2-

+

Case (4)

Case (5)

Case (6)

P

PbL

S.F.D

B.M.D

PabL

-+

PaL

Pb/L

Pa/L

7t

S.F.D

B.M.D

-

+

7t 7t

7t

7t

7t

B.M.D

7*2=14t.m

P

S.F.D

B.M.D

-

+

P P

P

P

P

B.M.D

P*a P*a


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1t

10t.m

1t

S.F.D-

1t

5t.m

B.M.D

@M@

PART (4)

2014

2014

4/6


Case (7)

5t

S.F.D

B.M.D

5*93 =15t.m

-+

5t 5t

5t

5t

5t

P

S.F.D

B.M.D

PL3

-+

P P

P

P

P

PL3

Case (8)

Case (9)

2t

12t.m

S.F.D-

2t

2*126 =4t.m

B.M.D

2t

4*126 =8t.m

ML

M

S.F.D-

ML

MaL

MbL

B.M.D

ML

PL

M

PL

S.F.D-

M2

M2

B.M.D

ML

5t.m


8/15

@M@

PART (4)

2014

2014

4/7


S.F.D

B.M.D

wL2

+

wt/m`

wL

wL8

Case (10)

S.F.D

B.M.D

2*52=25t.m

+

2t/m`

2*5 =10t

wL8

Case (11)

2*82=8t

S.F.D

B.M.D

-

+

2t/m`

8t

8t

8t

16t.m

16t.m

wL2

S.F.D

B.M.D

-

+

wt/m`

wL2

wL2

wL2

wL8

wL8

S.F.D

#

-1

-2B.M.D

-3Maximum MomentZero Shear Q=0t

Parabola

2

1

3

w

wL16t

Distributed loads


9/15

@M@

PART (4)

2014

2014

4/8


SOLVED EXAMPLES

Ex(2)

Ya

Xa0t

3t/m`

12t

6t

6t

Yb12t

0.0t N.F.D

M b= 0.0

Ya(6) - 12(4) + 6(2) = 0.0

FX= 0.0 Xa= 0.0 t

Ya= 6 t

REACTIONS

FY= 0.0 Yb= 12t

-

+ +S.F.D

6t

6t

6t12t

B.M.D

WL8=6m.t

12t.m

6t

6t 8t.m 5t4

3 3t

4t

Ex(1)

M b= 0.0

Ya(6) - 6(4) + +8 + 4(2) = 0.0

FX= 0.0 Xa= 3t

Ya= 1.33t

REACTIONS

FY= 0.0 Yb= 8.67t

Ya

Xa3t

1.33t 8.67t Yb

3t

N.F.D

1.33t

4.67t

4t S.F.D8.67t

6t

2.66t.m

6.68t.m

1.32t.m

8t.m

B.M.D

8t.m

2nd Par.

+

-

-

0t.m

Required Draw Internal Forces Diagrams (N.F.D , S.F.D & B.M.D).


10/15

@M@

PART (4)

2014

2014

4/9


0.0t

3t/m`4t 4t

18t

FX= 0.0 Xa= 0t

Ya13t Yb13t

Xa0tREACTIONS

LOAD symmetry

Ya= Yb=18+4*2

2 = 13.0t N.F.D

S.F.D+ +

4t

9t4t

9t

WL8=13.5m.t

2nd Par.

Ex(3)

WL8=13.5m.t

--

B.M.D

8t.m 8t.m

4t

4t/m`

12t

12t

4t/m`

6t 6t

4t

6t 6t

10t

2t 2t

4t

- -

++

S.F.D

B.M.D

12t.m

12t.m

WL8=4.5m.t

WL8=4.5m.t

Ya

Xa

Yb

4t/m`

12t

12t

4t/m`

6t 6tEx(4)

M b= 0.0

Ya(6) + 6(8)-12 (4.5)+12(1.5)

FX= 0.0 Xa= 0.0 t

Ya= - 4t()

REACTIONS

FY= 0.0 Yb= 4t

+6(2) = 0.0

Anti-Symmetric load

Symmetric S.F.D

Anti-Symmetric B.M.D

Symmetric load

Anti-Symmetric

Symmetric B.M.D


11/15

@M@

PART (4)

2014

2014

4/10


2nd Par.

PL4=6m.t

WL/8=24m.t

++

-+

-

10t

43

8t

6t3t/m`

Ya

Xa8t

Yb19t 23t

24t

6t6t

6t.m

Ex(5)

M b= 0.0

-Ya(12) + 6(14)+24 (8)+6(2) -6-6(1) =0

FX= 0.0 Xa= 8t

Ya= 23t

REACTIONS

FY= 0.0 Yb= (6+6+24+6)-23 = 19t

CHECK

M a= -6-6*13+19*12-6*10-24*4+6*2 =0 .......O.K.

10t

43

8t6t3t/m`

Ya

Xa8t

Yb19t 23t

24t

6t6t

6t.m

6t

13t

7t 6t

8t

17t

6t.m

12t.m

14t.m

12t.m

28t.m

23t

19t6t

S.F.D

B.M.D

N.F.D0.0t

INTERNAL FORCES


12/15

2nd Par.

WL8=1m.t

++

-

@M@

PART (4)

2014

2014

4/11


Ex(6)

Yb 25t

5t10t2t/m`

Ya

Xa0t

10t

10t.m

M b= 0.0

Ya(8) - 10 -10*2 - 2*10 (3) + 5(2) =0

FX= 0.0 Xa= 0t

Ya= 10t

REACTIONS

FY= 0.0 Yb= (20+10+5)-10 = 25tCHECK

M a= 5*10-25*8+10*6+20*5-10 = 0 .......O.K.

4t 4t 4t

20t

25t

5t10t2t/m`

10t

10t.m

12t 4t 4t 4t

10t

2t

12t16t

9t5t

14t.m

24t.m

14t.m14t.m

WL8=4m.t

S.F.D

B.M.D

25t

INTERNAL FORCES

8t

10t.m

24t.m


13/15

2nd Par.

+

- - -

--

N.F.D

S.F.D

B.M.D

B

@M@

PART (4)

2014

2014

4/12


Ex(7)

2t/m` 10t.m

11t

4t

5t

4t

45

15t

15t

RA

8t

4t

4t

B2t/m` 10t.m

11t

4t

5t

4t

15t

15t 8t

4t

4t

15t 15t

4t

4t

5t

1t

11t

4t

M b= 0.0

R cos45(8) - 4(10) - 8(6) - 10 -4(3) - 4(2) -4(1.5) = 0.0

FX= 0.0 Xb= 15-4 = 11 t FY= 0.0 Yb= (4+8+4+4)-15 = 5t

Check M a= -4*1.5-5*8+4*6+12*3-10-4*1 = 0.0 .....O.K.

R= 152 t

REACTIONS

4t.m

24t.m

14t.m

6t.m

6t.m6t.m

16t.m

WL8=4m.t WL8=1m.t

15t

4t

10t.m

INTERNAL FORCES

6t.m

6t.m

M @ j= 0


14/15

2nd Par.

WL/8=16m.t

+ +-

-

-

N.F.D

S.F.D

B.M.D

++

@M@

PART (4)

2014

2014

4/13


Ex(8)

b

Yb

Xb

5t

3

4

3t

4t

3

45

cos = 35

sin = 45R

Rsin45

Rcos45

45

6t

16t

2t/m`

M b= 0.0

R cos45(8) - 4(10) - 16(4) + 6(1) = 0.0

FX= 0.0 Xb= 12.25+3 = 15.25 t

FY= 0.0 Yb= (6+16+4) - 12.25 = 13.75 t

Check M a= 6(9) - 13.75(8) + 16(4) - 4(2) = 0.0

a

12.25t13.75t

15.25t

R= 12.252 t

b3t

4t

45

6t

16t

2t/m`a

12.25t 13.75t

15.25t12.25t

3t

15.25t6t

6t

6t

7.75t

8.25t

4t

6t.m

6t.m

6t.m8t.m

REACTIONS

INTERNAL FORCES

6t.m

6t.m

6t.m

6t.m

6t.m

6t


15/15

@M@

PART (4)

2014

2014


Ex(9)2t8t

4t/m`16t16t

8t

= 8.0 tX B

YA = 20 t

- 4 -16(8) - 10(4) -8(1) -16(2) +8(1) + 2(2) = 0Y (10)A

Check

Draw N.F.D , S.F.D & B.M.DA

2nd Parabola

10t 8t

20t

4t

6t

22t

10t

2t

4t.m

44t.m

12t.m

44t.m

8t.m

52t.m

4t.m

2nd Parabola

2nd Parabola

WL8=8t.m

N.F.D

S.F.D

B.M.D

YB = (10+16+16+8+2)-20 = 32 t

20t

10t

2t8t

20t 32t

16t 16t 8t

8t

4t.m

YA

XB

YB

2012/2013Mid-Term

Exam

REACTIONS

INTERNAL FORCES

WL8=8t.m

10t

8t

+

+

-

+

-

-

4t/m` 4t/m`

FX= 0.0

M B = 0.0

FY= 0.0

M A = 2*12+24*9-32*10-8*1+10*6+16*2-4 =0.0 O.K.

44t.m52t.m

8t.m

4t/m`

#END PART (4)

4-internal forces beams _1 model (1)

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